FOUND.OF COLLEGE CHEMISTRY
FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
Question
Book Icon
Chapter 14, Problem 22PE

(a)

Interpretation Introduction

Interpretation:

Grams of solute in 1.20 L of 18 M H2SO4 have to be determined.

Concept Introduction:

Molarity is amount of solute per 1 L of solution. Formula for molarity is as follows:

  Molarity of solution=Moles of soluteVolume (L) of solution

(a)

Expert Solution
Check Mark

Answer to Problem 22PE

2118.5 g is present in 1.20 L of 18 M H2SO4.

Explanation of Solution

Formula for molarity of H2SO4 solution is as follows:

  Molarity of H2SO4 solution=Moles of H2SO4Volume (L) of H2SO4 solution        (1)

Rearrange equation (1) for moles of H2SO4.

  Moles of H2SO4=[(Molarity of H2SO4 solution)(Volume (L) of H2SO4 solution)]        (2)

Substitute 18 M for molarity and 1.20 L for volume of H2SO4 solution in equation (2).

  Moles of H2SO4=(18 M)(1.20 L)=21.6 mol

Formula for mass of H2SO4 is as follows:

  Mass of H2SO4=(Moles of H2SO4)(Molar mass of H2SO4)        (3)

Substitute 21.6 mol for moles and 98.079 g/mol for molar mass of H2SO4 in equation (3).

  Mass of H2SO4=(21.6 mol)(98.079 g/mol)=2118.5 g

Hence, 2118.5 g is present in 1.20 L of 18 M H2SO4.

(b)

Interpretation Introduction

Interpretation:

Grams of solute in 27.5 mL of 1.50 M KMnO4 have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 22PE

6.52 g is present in 27.5 mL of 1.50 M KMnO4.

Explanation of Solution

Formula for molarity of KMnO4 solution is as follows:

  Molarity of KMnO4 solution=Moles of KMnO4Volume (L) of KMnO4 solution        (4)

Rearrange equation (4) for moles of KMnO4.

  Moles of KMnO4=[(Molarity of KMnO4 solution)(Volume (L) of KMnO4 solution)]        (5)

Substitute 1.50 M for molarity and 27.5 mL for volume of KMnO4 solution in equation (5).

  Moles of KMnO4=(1.50 M)(27.5 mL)(103 L1 mL)=0.04125 mol

Formula for mass of KMnO4 is as follows:

  Mass of KMnO4=[(Moles of KMnO4)(Molar mass of KMnO4)]        (6)

Substitute 0.04125 mol for moles and 158.034 g/mol for molar mass of KMnO4 in equation (6).

  Mass of KMnO4=(0.04125 mol)(158.034 g/mol)=6.52 g

Hence, 6.52 g is present in 27.5 mL of 1.50 M KMnO4.

(c)

Interpretation Introduction

Interpretation:

Grams of solute in 120 mL of 0.025 M Fe2(SO4)3 have to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 22PE

1.20 g is present in 120 mL of 0.025 M Fe2(SO4)3.

Explanation of Solution

Formula for molarity of Fe2(SO4)3 solution is as follows:

  Molarity of Fe2(SO4)3 solution=Moles of Fe2(SO4)3Volume (L) of Fe2(SO4)3 solution        (7)

Rearrange equation (7) for moles of Fe2(SO4)3.

  Moles of Fe2(SO4)3=[(Molarity of Fe2(SO4)3 solution)(Volume (L) of Fe2(SO4)3 solution)]        (8)

Substitute 0.025 M for molarity and 120 mL for volume of Fe2(SO4)3 solution in equation (8).

  Moles of Fe2(SO4)3=(0.025 M)(120 mL)(103 L1 mL)=0.003 mol

Formula for mass of Fe2(SO4)3 is as follows:

  Mass of Fe2(SO4)3=[(Moles of Fe2(SO4)3)(Molar mass of Fe2(SO4)3)]        (9)

Substitute 0.003 mol for moles and 399.88 g/mol for molar mass of Fe2(SO4)3 in equation (9).

  Mass of Fe2(SO4)3=(0.003 mol)(399.88 g/mol)=1.20 g

Hence, 1.20 g is present in 120 mL of 0.025 M Fe2(SO4)3.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A solution of 14 g of a nonvolatile, nonelectrolyte compound in 0.10 kg of benzene boils at 81.7°C. If the BP of pure benzene is 80.2°C and the K, of benzene is 2.53°C/m, calculate the molar mass of the unknown compound. AT₁ = Km (14)
Please help me answer the following questions. My answers weren't good enough. Need to know whyy the following chemicals were not used in this experiment related to the melting points and kf values. For lab notebook not a graded assignments.
Draw the arrow pushing reaction mechanism. DO NOT ANSWER IF YOU WONT DRAW IT. Do not use chat gpt.

Chapter 14 Solutions

FOUND.OF COLLEGE CHEMISTRY

Ch. 14.5 - Prob. 14.11PCh. 14.5 - Prob. 14.12PCh. 14 - Prob. 1RQCh. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQCh. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 18RQCh. 14 - Prob. 19RQCh. 14 - Prob. 20RQCh. 14 - Prob. 21RQCh. 14 - Prob. 22RQCh. 14 - Prob. 23RQCh. 14 - Prob. 24RQCh. 14 - Prob. 25RQCh. 14 - Prob. 26RQCh. 14 - Prob. 27RQCh. 14 - Prob. 28RQCh. 14 - Prob. 29RQCh. 14 - Prob. 30RQCh. 14 - Prob. 31RQCh. 14 - Prob. 32RQCh. 14 - Prob. 33RQCh. 14 - Prob. 34RQCh. 14 - Prob. 35RQCh. 14 - Prob. 37RQCh. 14 - Prob. 38RQCh. 14 - Prob. 39RQCh. 14 - Prob. 40RQCh. 14 - Prob. 41RQCh. 14 - Prob. 42RQCh. 14 - Prob. 1PECh. 14 - Prob. 2PECh. 14 - Prob. 3PECh. 14 - Prob. 4PECh. 14 - Prob. 5PECh. 14 - Prob. 6PECh. 14 - Prob. 7PECh. 14 - Prob. 8PECh. 14 - Prob. 9PECh. 14 - Prob. 10PECh. 14 - Prob. 11PECh. 14 - Prob. 12PECh. 14 - Prob. 13PECh. 14 - Prob. 14PECh. 14 - Prob. 15PECh. 14 - Prob. 16PECh. 14 - Prob. 17PECh. 14 - Prob. 18PECh. 14 - Prob. 19PECh. 14 - Prob. 20PECh. 14 - Prob. 21PECh. 14 - Prob. 22PECh. 14 - Prob. 23PECh. 14 - Prob. 24PECh. 14 - Prob. 25PECh. 14 - Prob. 26PECh. 14 - Prob. 27PECh. 14 - Prob. 28PECh. 14 - Prob. 29PECh. 14 - Prob. 30PECh. 14 - Prob. 31PECh. 14 - Prob. 32PECh. 14 - Prob. 33PECh. 14 - Prob. 34PECh. 14 - Prob. 35PECh. 14 - Prob. 36PECh. 14 - Prob. 37PECh. 14 - Prob. 38PECh. 14 - Prob. 39PECh. 14 - Prob. 40PECh. 14 - Prob. 41PECh. 14 - Prob. 42PECh. 14 - Prob. 44PECh. 14 - Prob. 45PECh. 14 - Prob. 46PECh. 14 - Prob. 47PECh. 14 - Prob. 48PECh. 14 - Prob. 49PECh. 14 - Prob. 50PECh. 14 - Prob. 51PECh. 14 - Prob. 52PECh. 14 - Prob. 53AECh. 14 - Prob. 54AECh. 14 - Prob. 55AECh. 14 - Prob. 56AECh. 14 - Prob. 57AECh. 14 - Prob. 58AECh. 14 - Prob. 59AECh. 14 - Prob. 60AECh. 14 - Prob. 61AECh. 14 - Prob. 62AECh. 14 - Prob. 63AECh. 14 - Prob. 65AECh. 14 - Prob. 66AECh. 14 - Prob. 67AECh. 14 - Prob. 68AECh. 14 - Prob. 69AECh. 14 - Prob. 70AECh. 14 - Prob. 71AECh. 14 - Prob. 72AECh. 14 - Prob. 73AECh. 14 - Prob. 74AECh. 14 - Prob. 75AECh. 14 - Prob. 76AECh. 14 - Prob. 77AECh. 14 - Prob. 78AECh. 14 - Prob. 79AECh. 14 - Prob. 80AECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Prob. 83AECh. 14 - Prob. 84AECh. 14 - Prob. 85AECh. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - Prob. 88AECh. 14 - Prob. 90AECh. 14 - Prob. 91AECh. 14 - Prob. 92AECh. 14 - Prob. 93AECh. 14 - Prob. 94AECh. 14 - Prob. 95AECh. 14 - Prob. 96AECh. 14 - Prob. 97AECh. 14 - Prob. 98AECh. 14 - Prob. 99CECh. 14 - Prob. 100CECh. 14 - Prob. 102CECh. 14 - Prob. 103CECh. 14 - Prob. 104CECh. 14 - Prob. 105CE
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Fundamentals Of Analytical Chemistry
Chemistry
ISBN:9781285640686
Author:Skoog
Publisher:Cengage